C. Help Shahhoud

Shahhoud is participating in the first Div.3 contest on Codeforces, the first problem was:

Given two strings A and B of equal length N (N is odd), find the minimum number of steps needed to change A into B, or print  - 1 if it’s impossible.

In each step, you can choose an odd integer x such that 1 ≤ x ≤ N, and reverse the substring of length x that is centered in the middle of A. For example, performing a step with x = 3 on the string “abcde” results in “adcbe” and applying x = 5 on “abcde” results in “edcba”.

Can you help Shahhoud solve the problem?

Input

The first line contains one integer T, the number of test cases.

Each test case consists of two lines, the first contains the string A, and the second contains the string B. (1 ≤ |A| = |B| ≤ 105) |A| = |B| is odd.

Both strings consist of lowercase English letters.

Output

For each test case, print one line containing one integer,  - 1 if A can’t be changed into B, or the minimum number of steps to change A into B.

Example

Input

1
abcxdef
fecxdba

Output

2

题意

给两个一样长的，长度为n(n%2==1)n(n \% 2==1)奇数的串，每一次都以(n+1)/2(n+1)/2为中心翻转，每一次翻转的长度是自己决定的，即每次操作都是翻转一个x(1<=x<(n+1)/2)x(1<=x<(n+1)/2)为起点，n−x+1n-x+1为终点的串，e.g.

if(x=2),abcde==>adcbeif(x=2),\\ abcde==>adcbe

怎样用最少的翻转次数使两个串相同，输出最少的次数，如果不能使两个串相同输出-11

解

翻转策略一定从外往里翻转最优，复杂度O(N)
翻转的时候不要真的在字符串上操作，不要去swap，只需要一个变量time记录已经翻了多少次，当time为偶数的时候，当前子串就是原来的样子，time为奇数的时候当前字串是已经翻转过一次的
e.g.

设S=abcdefg当time%2==1,i==3S没有变，但串实际上是S′=abedcfg,我们就按照S′来做当time%2==0,i==3串实际上为S′=abcdefg设S=abcdefg\\ 当time \% 2 == 1,i==3\\ S没有变，但串实际上是S'=abedcfg,我们就按照S'来做 \\ 当time\% 2==0,i==3\\ 串实际上为S'=abcdefg 

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string a, b;
int time;
int len;
int T;
void rev(int t) {
	time++;
}
int main()
{
	scanf("%d", &T);
	while (T--)
	{
		time = 0;
		cin >> a >> b;
		len = a.size();
		int flag = 0;
		if (a[(len) / 2] != b[(len) / 2])//中间的不一样
		{
			printf("-1\n");
			continue;
		}
		for (int i = 0; i < (len) / 2; i++) {
			if (a[i] != b[i] && a[i] != b[len - 1 - i])//翻不翻都不能一样
			{
				flag = 1;
				break;
			}
			if ((time & 1) == 0 && a[i] != b[i]) {
				if (a[i] == b[len - 1 - i] && b[i] == a[len - 1 - i])
				{
					rev(i);
				}
			}
			else if ((time & 1) == 1 && a[i] != b[len - 1 - i]) {
				if (a[i] == b[i] && b[len - 1 - i] == a[len - 1 - i])
				{
					rev(i);
				}
			}
		}
		if (flag == 1)printf("-1\n");
		else printf("%d\n", time);
	}
}